Question

If $u=\log \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$ , then prove that $\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}\right)\left(\dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=1$

Answer 1

Hint: In the question, first work out on differentiation of the given function u two times with respective to x, y and z respectively. Second work out on the left hand side of the given expression and their simplification.

Complete step-by-step answer:
 It is given that
$u=\log \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
$u=\log {{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{\dfrac{1}{2}}}$
By using the logarithmic rule, we get
$u=\dfrac{1}{2}\log \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$
Multiplying both sides by 2, we get
$2u=\log \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)................(1)$
Differentiating with respect to x on both sides, we get
$2\dfrac{du}{dx}=\dfrac{1}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}(2x)$
Dividing both sides by 2, we get
$\dfrac{du}{dx}=\dfrac{x}{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}..............(2)$
Differentiating the equation (2) with respect to x by using quotient rule $\left( \dfrac{d}{dx}\dfrac{u}{v} \right)=\dfrac{v\left( \dfrac{du}{dx} \right)-u\left( \dfrac{dv}{dx} \right)}{{{v}^{2}}}$ , we get
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{dx}{dx} \right)-x\left( \dfrac{d}{dx}({{x}^{2}}+{{y}^{2}}+{{z}^{2}}) \right)}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}$
Rearranging the terms, we get
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)-x\left( 2x \right)}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}=\dfrac{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2{{x}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}$
Finally we get
$\dfrac{{{d}^{2}}u}{d{{x}^{2}}}=\dfrac{-{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}$
Similarly, differentiate the given function with respect to y and z , we get
$\dfrac{{{d}^{2}}u}{d{{y}^{2}}}=\dfrac{{{x}^{2}}-{{y}^{2}}+{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}$ and $\dfrac{{{d}^{2}}u}{d{{z}^{2}}}=\dfrac{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}$
Let us consider the left hand side,
$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left[ \dfrac{-{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+{{x}^{2}}-{{y}^{2}}+{{z}^{2}}+{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} \right]$
Cancelling the like terms on the right side, we get
$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left[ \dfrac{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} \right]$
Rearranging the terms, we get
$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=\left[ \dfrac{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}}{{{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{2}}} \right]$
Cancelling the terms on the right side, we get
$\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)\left( \dfrac{{{d}^{2}}u}{d{{x}^{2}}}+\dfrac{{{d}^{2}}u}{d{{y}^{2}}}+\dfrac{{{d}^{2}}u}{d{{z}^{2}}} \right)=1$
This is a desired result.

Note:We might get confused about the difference between the ordinary differentiation and partial differentiation. In ordinary differentiation, we find derivatives with respect to one variable only, as function contains only one variable. Partial differentiation is used to differentiate mathematical functions having more than one variable in them.